The optimal path of Liss is as follows: First she starts from the root of tree 1. Walk up the tree to the top and eat a nut. Walk down to the height min(h1, h2). Jump to the tree 2. Walk up the tree to the top and eat a nut. Walk down to the height min(h2, h3), ... and so on.

In this problem, there are many simple algorithms which works in O(n). One of them (which I intended) is following:

You should prepare 2 vectors. If s[i] = 'l', you should push i to the first vector, and if s[i] = 'r', you should push i to the second vector. Finally, you should print the integers in the second vector by default order, after that, you should print the integers in the first vector in the reverse order.

This algorithm works because if Liss divides an interval into two intervals A and B and she enters A, she will never enter B.

The main idea is DP. Let's define dp[x] as the maximal value of the length of the good sequence whose last element is x, and define d[i] as the (maximal value of dp[x] where x is divisible by i).

You should calculate dp[x] in the increasing order of x. The value of dp[x] is (maximal value of d[i] where i is a divisor of x) + 1. After you calculate dp[x], for each divisor i of x, you should update d[i] too.

This algorithm works in O(nlogn) because the sum of the number of the divisor from 1 to n is O(nlogn).

Note that there is a corner case. When the set is {1}, you should output 1.

There are many O(Q * N * logN) solutions using segment trees or other data structures, but probably they will get time limit exceeded.

We can solve each query independently. First, let's consider the following DP algorithm.

dp[c] := the maximal value of a sequence whose last ball's color is c

For each ball i, we want to update the array. Let the i-th ball's color be col[i], the i-th ball's value be val[i], and the maximal value of dp array other than dp[col[i]] be otherMAX. We can update the value of dp[col[i]] to dp[col[i]] + val[i] × a or otherMAX + val[i] × b. Here, we only need to know dp[col[i]] and otherMAX. If we remember the biggest two values of dp array in that time and their indexes in the array, otherMAX can be calculated using the biggest two values, which always include maximal values of dp array other than any particular color.

Since the values of dp array don't decrease, we can update the biggest two values in O(1). Finally, the answer for the query is the maximal value of dp array.

First, let's consider a simpler version of the problem: You are given a start state and a goal state. Check whether the goal state is reachable from the start state.

Define A, B, C, and D as in the picture below, and let I be the string of your instructions. A and B are substrings of s, and C and D are substrings of t.

It is possible to reach the goal state from the start state if there exists an instruction I such that:

1 A is a subsequence of I.

2 B is not a subsequence of I.

3 C is a subsequence of I.

4 D is not a subsequence of I.

So we want to check if such string I exists. (string s1 is called a subsequence of s2 if it is possible to get s2 by removing some characters of s1)

There are some obvious "NO" cases. When D is a subsequence of A, it is impossible to satisfy both conditions 1 and 4. Similarly, B must not be a subsequence of C. Are these sufficient conditions? Let's try to prove this hypothesis.

To simplify the description we will introduce some new variables. Let A', B', C', and D' be strings that can be obtained by removing the first characters of A, B, C, and D. Let c1 and c2 be the first characters of A and C.

Suppose that currently the conditions are satisfied (i.e. D is not a subsequence of A and B is not a subsequence of C).

If c1 = c2, you should perform the instruction c1 = c2. The new quatruplet will be (A', B', C', D') and this also satisies the conditions.

If c1 ≠ c2 and B' is not a subsequnce of C, you should perform the instruction c1. The new quatruplet will be (A', B', C, D) and this also satisies the conditions.

If c1 ≠ c2 and D' is not a subsequnce of A, you should perform the instruction c2. The new quatruplet will be (A, B, C', D') and this also satisies the conditions.

What happens if all of the above three conditions don't hold? In this case A and C have the same length and A = c1c2c1c2..., B = c2c1c2c1. In particular the last two characters of A and B are swapped: there are different characters x and y and A = ...xy, B = ...yx. Now you found a new necessary condition! Generally, if A and B are of the form A = ...xy and B = ...yx, the goal state is unreachable. If the last instruction is x, Vasya must be in the goal before the last instruction, but then Vasya will go further after the last instruction. If the last instruction is y, we will also get a contradiction.

Finally we have a solution. The goal state is reachable from the start state if and only if D is not a subsequence of A, B is not a subsequnce of C, and A and C are not of the form A = ...xy, C = ...yx. The remaining part is relatively easy, so I'll leave it as an exercise for readers.

Thank you. Tommorow there will be a match, and now I'm deciding which language to use. I have looked through all the submissions to that tasks, way most of them are written with C++, some Java, and even two Python solutions. Nothing else. My "second favourite" language is C#, and I have written a solution to the same problem in C#, but... Превышено ограничение времени на тесте 35 (TLE on 35th). So, C# is too slow too? Here is my C# code

In one of Java solutions I've seen class "FastScanner" with only method "FastRead", which reads from STDIO. If you think that the time issue is in IO, maybe there is a possibility to write some wrappers aroung Ruby's gets and puts?

I have tried same in Ruby, and now it fails on 45th, not 31st test. I just replaced that puts left with puts left.join("\n"), and the perfomance went up dramaticly. So, it seems that the weakest part in Ruby is the IO. Will think about it more.

because "You should calculate dp[x] in the increasing order of x" and we are using divisors not all integers that have a common divisor with x ... you have also to imagine that for each new x , we are searching for the highest d [ i ] and inserting this x in all possible sequences ... that 's why "update" is necessary

I don't think editorial nor comments explanation on solution 264A - Escape from Stones are intuitive so I'll try to explain how I came up with a solution.

You can visualize the problem as a binary tree. Each node is responsable for a interval [a, b]. Starting from the root, responsable for [0, 1], if you go left then root will have a left child responsable for interval [0, 1 / 2]. Since you're a left child, last time the stone fell on position 1 / 2, the end point of the interval.

Note that from that child, it doesn't matter if you go left or right: the stone will never fall on a position bigger than 1 / 2 (the end point). The ideia is: since a stone always fall on the midpoint (a + b) / 2 of the lower and end points, it can never fall after the endpoint. In fact, the stone must fall infinitely many times to the right to reach that endpoint. So, we were on a node [0, 1], the stone fell and we jumped to the left, creating a left child [0, 1 / 2] and we now know that there is no way the stone will every fall to a position > 1 / 2. This means that in the final solution, the current node [0, 1 / 2] will come AFTER all nodes (descendents) that come after this one.

If we jumped to the right instead, creating a right child responsable for interval [1 / 2, 1] we can show the same way that the stone will never fall before the starting point 1 / 2. This means that the current node will com BEFORE all nodes (descendents) that come after this one, since we know that the stone can't fall on a postion < 1 / 2 from now on.

This is an algorithm: if in the i'th stone we jumped to the left, we now that stone i will become AFTER all next stones. If we jumped right, stone i will become BEFORE all next stones.

Hi akhi29Two points-1. Try to use StringBuilder (popular practice among Java user's) instead of directly printing to the output console.2. When you call, close() of PrintStream, it isn't obliged to flush your output to the console. You need to call flush() explicitly to ensure that the data, if present in the buffer is flushed to the output console. You can refer the modified submission 30322735

In DIV 2 C I took 2 variables hi = 1 and lo = 0, if character is 'l' then I just change hi = (hi + lo) / 2 otherwise lo = (hi + lo) / 2 and mapped the value, sort them and print them, but got WA on Testcase 11, could you please tell me what is I'm missing?

In Div 1 A Stones falls in 'k' position and print the stone position from left

I calculated 'k' and 'd' as k=(X+Y)/2 and d=(Y-X)/2 And if 'l' then x-=d otherwise y+=d. then sorted the array which contains 'k' s value and its position no. but my implementation isn't matching with the test case..